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Simple derivative. Where is the logical fallacy here?

  1. Dec 6, 2011 #1
    [itex]2^{2} = 2 + 2[/itex]
    [itex]3^{2} = 3 + 3 + 3[/itex]
    [itex]4^{2} = 4 + 4 + 4 + 4[/itex]

    Generalizing...
    [itex]x^{2} = x + ... + x[/itex] (x times)

    Take the derivative of both sides.
    [itex]\frac{\delta}{\delta x} x^{2}[/itex] = [itex]\frac{\delta}{\delta x}[ x + ... + x][/itex] (x times)
    [itex]2x = 1 + ... + 1[/itex] (x times) [itex]= x [/itex] !?

    Surely, there must be something wrong... and there is! But I thought this was pretty cool.
     
  2. jcsd
  3. Dec 6, 2011 #2

    This is true if x is a natural number only.
    In order to do derivative you need a real variable x and a function for which differentiation makes sense.You should really try to understand what are the conditions for differentiation to avoid this kind of nonsense.
     
  4. Dec 7, 2011 #3

    I like Serena

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    The derivative of your sequence fails to take into account that an extra term pops up.

    Look at it like this:
    [tex]\frac{d}{dx}f(x) \approx {f(x+1)-f(x) \over 1}[/tex]
    [tex]\frac{d}{dx}[x + ... + x] \approx {[(x+1) + ... + (x+1) + (x+1)] - [x + ... + x] \over 1} = 1 + ... + 1 + (x+1) = 2x+1[/tex]


    Note that it normal to approximate (squeeze) integrals with summations, and summations with integrals.
    Some major proofs are built on that concept.
     
  5. Dec 7, 2011 #4

    HallsofIvy

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    When you write "[itex]x^2= x+ x+ \cdot\cdot\cdot+ x[/itex], x times", the "x times" summation is itself a function of x. You cannot just differentiate the sum as if the number of terms were a constant.
     
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